If one root of the equation $x^{2} - \lambda x + 12 = 0\ is\ even\ prime\ while\ x^{2} + \lambda x + \mu = 0$ has equal roots, then $\mu$ is

Question

Dr. Abubacker · Accepted Answer

(b)
We know that only even prime is 2, then
$(2)^{2} - \lambda(2) + 12 = 0\  \Rightarrow \ \lambda = 8$ …(i)
and $x^{2} + \lambda x + \mu = 0$ has equal roots.
$	herefore\ \lambda^{2} - 4\mu = 0$ or $(8)^{2} - 4\mu = 0$ [ from Eq. (i)]
$	herefore\ \mu = 16$

Q. If one root of the equation $x^{2} - \lambda x + 12 = 0\ is\ even\ prime\ while\ x^{2} + \lambda x + \mu = 0$ has equal roots, then $\mu$ is

Explanation

Reviewed by Dr. Abubacker

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